Suppose to have a Lie algebra L with a reductive lie subalgebra G. Let l an element of L such that [l,g] is in G for every g in G, is it true that l is an element of G?if not, there are some restriction on G that makes it true?

2 Answers
2

If $G$ is semisimple, then every derivation of $G$ is inner, so
that the normalizer $N_L(G)=C_L(G)+G$ where $C_L(G)$
is the centralizer. In this situtaion Michele's condition holds
if and only if the centralizer is trivial.

However the case where $G$ is reductive isn't so easy.
A reductive $G$ is the direct sum of an Abelian and a semisimple
Lie algebra. The Abelian part can have (if at least two-dimensional)
non-trivial derivations. These will lift to non-inner derivations
of $G$. If we let $L$ be the semidirect product of $G$ with
a one-dimensional Lie algebra using this derivation, then
$L$ normalizes $G$, $L\ne G$ but $C_L(G)$ is trivial.

The set of $l$ such that $[l,g]\in G$ for all $G$ is called the normalizer of $G$. It's not particularly common for the normalizer to be the same as G, though it does happen sometimes. There are a few theorems I know about specific Lie algebras being self-normalizing, such as Cartan and Borel subalgebras in reductive Lie algebras, but it's not very common and I don't know any general condition which guarantees it.